# Constant Acceleration of Two Cars

• ff4930
In summary, two cars are traveling along a straight road, with car A maintaining a constant speed of 80 km/h and car B maintaining a constant speed of 110 km/h. At t=0, car B is 45 km behind car A. To find how much farther car A will travel before car B overtakes it, an equation of 110x = 80x+45 is used, where x represents time. Solving for x, it is found that car A will travel another 120 km before car B catches up. To find how much ahead of car A car B will be in 30 seconds after it overtakes it, the distance traveled by each car in 30 seconds is calculated. It is found
ff4930
[SOLVED] Constant Acceleration

## Homework Statement

Two cars are traveling along a straight road. Car A maintains a constant speed of 80 km/h and car B maintains a constant speed of 110 km/hr. At t=0, car B is 45 km behind Car A.
A) How much farther will car A travel before Car b overtakes it?
B)How much ahead of A will B be in 30 seconds after it overtakes A?

## The Attempt at a Solution

I am stumped. I read and reread the section on my textbook about constant acceleration and the material is so confusing. Can someone please hint me or point me to a direction of understanding this please. Thanks.

Why constant acceleration? Acceleration is 0 as you only have constant speed right?

Come up with expressions for the position of each car as a function of time. I'd call the location of Car b at t = 0 to be the origin. (So Car A is at 45 km at t = 0.)

There's no acceleration mentioned, so this should be easy.

110x = 80x+45
x=.5?
is this the right approach? I am really lost.

ff4930 said:
110x = 80x+45
x=.5?
is this the right approach? I am really lost.
Your equation is right, but how did you get x = .5? And what are the units? For that matter, what is x? What have you solved for here?

If you explain your reasoning a little more, we can help you better.

At the time instant where two cars meet up, the distance traveled by Car B will be 45 more than Car A...
v = x/t is the main equation for this type of question.

belliott4488 said:
Your equation is right, but how did you get x = .5? And what are the units? For that matter, what is x? What have you solved for here?

If you explain your reasoning a little more, we can help you better.
.5 hr? it will take .5 hr for Car B to catch up to Car A?

So Car A will travel another 40 km before Car B catches up.

Last edited:
Okay, so x is time (it's more conventional to use "t" for a time interval and "x" for distance, but it's not necessary). I think you just need to be a little more careful in solving your equation for x - you just made an algebra mistake.

Oops, thanks, Suppose to be 45/30 = 1.5.
So Car A will travel another 120 km before Car B catches up?

B)asks the distance Car B will be ahead of for 30 seconds after it passes Car A.
Car B 110 km/h(110 km per 3600 seconds), Car A 80 km/hr(80 km per 3600 seconds)
each second for Car B covers .031km and each second for Car A covers .022km
.031km * 30sec = .93 km.

So after 30 seconds, Car B will be .93 km ahead of Car A

Is this approach correct?

ff4930 said:
Oops, thanks, Suppose to be 45/30 = 1.5.
So Car A will travel another 120 km before Car B catches up?

B)asks the distance Car B will be ahead of for 30 seconds after it passes Car A.
Car B 110 km/h(110 km per 3600 seconds), Car A 80 km/hr(80 km per 3600 seconds)
each second for Car B covers .031km and each second for Car A covers .022km
.031km * 30sec = .93 km.

So after 30 seconds, Car B will be .93 km ahead of Car A

Is this approach correct?
Not quite - .031 km/sec * 30 sec gives the distance B has traveled, but A isn't just standing still - it's covered some distance, too. The difference between the two tells you how far B is ahead of A.

Cool! Thanks belliott, much obliged.

## 1. What is constant acceleration?

Constant acceleration is the rate at which an object's velocity changes over time, while its acceleration remains constant. This means that the object's speed will increase or decrease by the same amount every second.

## 2. How is constant acceleration calculated?

The formula for constant acceleration is a = (vf - vi) / t, where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time interval. This formula can be used to calculate the acceleration of two cars moving at a constant rate.

## 3. Can two cars have the same constant acceleration?

Yes, two cars can have the same constant acceleration if they both undergo the same change in velocity over the same time interval. This could occur if both cars have the same engine power and are driving on a flat surface with no external forces acting on them.

## 4. What is the relationship between acceleration and distance traveled?

The relationship between acceleration and distance traveled is described by the formula d = vi*t + 1/2*a*t^2, where d is the distance traveled, vi is the initial velocity, a is the acceleration, and t is the time interval. This means that as the acceleration increases, the distance traveled also increases.

## 5. How does air resistance affect the constant acceleration of two cars?

Air resistance, or drag, acts in the opposite direction of the motion of a car and can decrease its acceleration. This is because the force of air resistance pushes against the car, making it harder for the car to accelerate. Therefore, air resistance can affect the constant acceleration of two cars by potentially slowing them down.

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